?]]>2007-01-03T22:57:09ZFluxBBhttp://www.mathisfunforum.com/viewtopic.php?id=5592?]]>i have to graph it....can you help me on the graphing]]>http://www.mathisfunforum.com/profile.php?id=56912007-01-03T22:57:09Zhttp://www.mathisfunforum.com/viewtopic.php?pid=55202#p55202?]]>if a certain parabola has its vertex at (2,-1) and its focus at (2,1), write its equation and graph.
(h,k) = (2,-1) h = 2 k = -1 focus (h,k + p) is (2,1) so k + p = 1 k = -1 p = 2
y - k = 1/4p( x - h)^2 < equation y - (-1) = 1/4(2) (x - (2))^2 < substituted y = 1/8 ( x - 2)^2 - 1 < solved